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A bucket containing water is revolved in a vertical circle of radius ' $\mathrm{r}^{\prime}$. To prevent the water from falling down, the minimum frequency of revolution required is
$[\mathrm{g}=$ acceleration due to gravity $]$
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$[\mathrm{g}=$ acceleration due to gravity $]$
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Verified Answer
The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{r}}}$
For minimum velocity at the highest point we should have
$\mathrm{mr} \omega^{2}=\mathrm{mg}$
$\therefore \omega^{2}=\frac{g}{r} \quad$ or $\omega=\sqrt{\frac{g}{r}}$
$\begin{aligned} 2 \pi f &=\sqrt{\frac{g}{r}} \\ f &=\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \end{aligned}$
$\mathrm{mr} \omega^{2}=\mathrm{mg}$
$\therefore \omega^{2}=\frac{g}{r} \quad$ or $\omega=\sqrt{\frac{g}{r}}$
$\begin{aligned} 2 \pi f &=\sqrt{\frac{g}{r}} \\ f &=\frac{1}{2 \pi} \sqrt{\frac{g}{r}} \end{aligned}$
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